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MyWikiBiz, Author Your Legacy — Thursday May 02, 2024
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none o yer fuzzy grey square bullets fer me
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'''Definition.'''  Let the function <math>\lnot_j : \mathbb{B}^k \to \mathbb{B}</math> be defined for each integer <math>j\!</math> in the interval <math>[1, k]\!</math> by the following equation:
 
'''Definition.'''  Let the function <math>\lnot_j : \mathbb{B}^k \to \mathbb{B}</math> be defined for each integer <math>j\!</math> in the interval <math>[1, k]\!</math> by the following equation:
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| <math>\lnot_j (x_1, \ldots, x_j, \ldots, x_k) ~=~ x_1 \land \ldots \land x_{j-1} \land \lnot x_j \land x_{j+1} \land \ldots \land x_k.</math>
 
| <math>\lnot_j (x_1, \ldots, x_j, \ldots, x_k) ~=~ x_1 \land \ldots \land x_{j-1} \land \lnot x_j \land x_{j+1} \land \ldots \land x_k.</math>
 
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Then <math>\nu_k : \mathbb{B}^k \to \mathbb{B}</math> is defined by the following equation:
 
Then <math>\nu_k : \mathbb{B}^k \to \mathbb{B}</math> is defined by the following equation:
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{| align="center" cellpadding="8"
 
| <math>\nu_k (x_1, \ldots, x_k) ~=~ \lnot_1 (x_1, \ldots, x_k) \lor \ldots \lor \lnot_j (x_1, \ldots, x_k) \lor \ldots \lor \lnot_k (x_1, \ldots, x_k).</math>
 
| <math>\nu_k (x_1, \ldots, x_k) ~=~ \lnot_1 (x_1, \ldots, x_k) \lor \ldots \lor \lnot_j (x_1, \ldots, x_k) \lor \ldots \lor \lnot_k (x_1, \ldots, x_k).</math>
 
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For example, consider two cases at opposite vertices of the cube:
 
For example, consider two cases at opposite vertices of the cube:
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* The point <math>(1, 1, \ldots , 1, 1)</math> with all 1's as coordinates is the point where the conjunction of all posited variables evaluates to <math>1,\!</math> namely, the point where:
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:: <math>x_1 ~ x_2 ~\ldots~ x_{n-1} ~ x_n ~=~ 1.</math>
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| valign="top" | <big>&bull;</big>
 
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| The point <math>(1, 1, \ldots , 1, 1)</math> with all 1's as coordinates is the point where the conjunction of all posited variables evaluates to <math>1,\!</math> namely, the point where:
* The point <math>(0, 0, \ldots , 0, 0)</math> with all 0's as coordinates is the point where the conjunction of all negated variables evaluates to <math>1,\!</math> namely, the point where:
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:: <math>\texttt{(} x_1 \texttt{)(} x_2 \texttt{)} \ldots \texttt{(} x_{n-1} \texttt{)(} x_n \texttt{)} ~=~ 1.</math>
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| &nbsp;
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| align="center" | <math>x_1 ~ x_2 ~\ldots~ x_{n-1} ~ x_n ~=~ 1.</math>
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|-
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| valign="top" | <big>&bull;</big>
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| The point <math>(0, 0, \ldots , 0, 0)</math> with all 0's as coordinates is the point where the conjunction of all negated variables evaluates to <math>1,\!</math> namely, the point where:
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|-
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| &nbsp;
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| align="center" | <math>\texttt{(} x_1 \texttt{)(} x_2 \texttt{)} \ldots \texttt{(} x_{n-1} \texttt{)(} x_n \texttt{)} ~=~ 1.</math>
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To pass from these limiting examples to the general case, observe that a singular proposition <math>s : \mathbb{B}^k \to \mathbb{B}</math> can be given canonical expression as a conjunction of literals, <math>s = e_1 e_2 \ldots e_{k-1} e_k</math>.  Then the proposition <math>\nu (e_1, e_2, \ldots, e_{k-1}, e_k)</math> is <math>1\!</math> on the points adjacent to the point where <math>s\!</math> is <math>1,\!</math> and 0 everywhere else on the cube.
 
To pass from these limiting examples to the general case, observe that a singular proposition <math>s : \mathbb{B}^k \to \mathbb{B}</math> can be given canonical expression as a conjunction of literals, <math>s = e_1 e_2 \ldots e_{k-1} e_k</math>.  Then the proposition <math>\nu (e_1, e_2, \ldots, e_{k-1}, e_k)</math> is <math>1\!</math> on the points adjacent to the point where <math>s\!</math> is <math>1,\!</math> and 0 everywhere else on the cube.
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